Question

If one of the zeros of the quadratic polynomial $\left( k-1 \right){{x}^{2}}+kx+1$ is -3, then the value of k is
A.$\dfrac{4}{3}$
B.$-\dfrac{4}{3}$
C.$\dfrac{2}{3}$
D.$-\dfrac{2}{3}$

Answer 1

Hint:For solving this problem, first express the general quadratic equation in terms of sum of zeroes and product of zeros. Let the other zero be m. Now, we evaluate both the sum and product of zeroes. Hence, we have two variables (k and m) and two equations so we can obtain the desired result.

Complete Step-by-step answer:
In algebra, a quadratic function is a polynomial function with one or more variables in which the highest-degree term is of the second degree. A single-variable quadratic function can be stated as:
$f(x)=a{{x}^{2}}+bx+c,\quad a\ne 0$
In the above equation, the sum of zeroes is given as $\dfrac{-b}{a}$ and the product of zeroes is given as $\dfrac{c}{a}$.
According to the problem statement, we are given quadratic polynomials $\left( k-1 \right){{x}^{2}}+kx+1$ whose one root is -3. Let the other root be m. So, the sum of the roots can be expressed as:
$m-3=\dfrac{-k}{k-1}\ldots (1)$
The product of roots can also be expressed as:
$\begin{align}
  & -3m=\dfrac{1}{k-1} \\
 & m=\dfrac{-1}{3\left( k-1 \right)}\ldots (2) \\
\end{align}$
Now by using both equation we can solve for k. by putting the value of m from equation (2) to equation (1), we get
$\begin{align}
  & \dfrac{-1}{3k-3}-3=\dfrac{-k}{k-1} \\
 & \dfrac{-1-3\left( 3k-3 \right)}{3k-3}=\dfrac{-k}{k-1} \\
 & \dfrac{-1-9k+9}{3\left( k-1 \right)}=\dfrac{-k}{k-1} \\
 & \dfrac{8-9k}{3\left( k-1 \right)}=\dfrac{-k}{k-1} \\
 & 8-9k=\dfrac{-k}{k-1}\times 3\left( k-1 \right) \\
 & 8-9k=-3k \\
 & 9k-3k=8 \\
 & 6k=8 \\
 & k=\dfrac{8}{6} \\
 & k=\dfrac{4}{3} \\
\end{align}$
The value of k is $\dfrac{4}{3}$
Therefore, option (a) is correct.

Note:This problem can be alternatively solved by using the concept of zeroes of a polynomial. So, at x = -3, the value of f(x) = 0. Satisfying x = -3 in the given function, we get
$\begin{align}
  & \Rightarrow \left( k-1 \right){{\left( -3 \right)}^{2}}+k\left( -3 \right)+1=0 \\
 & 9k-9-3k+1=0 \\
 & k=\dfrac{4}{3} \\
\end{align}$
Therefore, by this methodology we obtain the same value of k.